Update Bits

题目：

Given two 32-bit numbers, N and M, and two bit positions, i and j. Write a method to set all bits between i and j in N equal to M (e g , M becomes a substring of N located at i and starting at j)

Given N=(10000000000)2, M=(10101)2, i=2, j=6
return N=(10001010100)2

分析：

这题目需要几个mask，但是如何形成mask需要注意，当j == 31和j != 31的时候要分开处理

背诵题，背过就是会，忘了就完蛋

解法：

class Solution {
public:
    /**
     *@param n, m: Two integer
     *@param i, j: Two bit positions
     *return: An integer
     */
    int updateBits(int n, int m, int i, int j) {
        // write your code here
        int max = ~0;
        if (j == 31) {
            j = max;
        } else {
            j = (1 << (j + 1)) - 1;
        }
        int left = max - j;
        int right = (1 << i) - 1;
        int mask = left | right;

        return (m << i) + (n & mask);
    }
};